从排列组合的角度考虑,把跳2阶当作黑球,跳1阶当作白球。问题就转化为,可以用多少个黑球以及黑球和白球的排列方法有几种。需要注意的是,用int会发生 浮点错误。

class Solution {
public:
    int jumpFloor(int number) {
        double sum = 0;
        int one;
        for(int two = 0; two <= number/2; two++){
            one = number - two * 2;
            sum += 1.0* factorial(one+two)/(factorial(one) * factorial(two));
        }
        return sum;
    }
    double factorial(int n){
        double r = 1;
        for(int i = 1; i <= n; i++)
            r *= i;
        return r;
    }
};